I'm putting this on the web because some students might
find it interesting. It could easily be mentioned in
many undergraduate math courses, though it doesn't seem
to appear in most textbooks used for those courses.
None of this material was discovered by me. --
ES

You should know that the solution of ax2+bx+c=0 is

There is an analogous formula for polynomials of degree
three: The solution of ax3+bx2+cx+d=0 is

(A formula like this was first published by Cardano in 1545.)
Or, more briefly,

x = {q + [q2
+ (r-p2)3]1/2}1/3
+ {q - [q2
+ (r-p2)3]1/2}1/3
+ p

where

p = -b/(3a), q = p3 + (bc-3ad)/(6a2), r = c/(3a)

But I do not recommend that you memorize these formulas.

Aside from the fact that it's too complicated, there
are other reasons why we don't teach this formula
to calculus students. One reason is that
we're trying to avoid teaching them about complex
numbers. Complex numbers (i.e., treating points
on the plane as numbers) are a more advanced topic,
best left for a more advanced course. But then the
only numbers we're allowed to use in calculus
are real numbers (i.e., the points on the line).
That imposes some restrictions on us --- for instance,
we can't take the square root of a negative
number. Now, Cardan's formula has the drawback
that it may bring such square roots into play
in intermediate steps of computation, even when those
numbers do not appear in the problem or its answer.

For instance, consider the cubic equation
x3-15x-4=0. (This example was
mentioned by Bombelli in his book in 1572.)
That problem has real
coefficients, and it has three real roots
for its answers. (Hint: One of the roots is
a small positive integer; now can you find all
three roots?)
But if we apply Cardano's formula to this example,
we use a=1, b=0, c=-15, d=-4, and we find that
we need to take the square root of -109 in
the resulting computation. Ultimately,
the square roots of negative numbers would cancel out
later in the computation, but that computation
can't be understood by a calculus student without
additional discussion of complex numbers.

There is also an analogous formula for polynomials of
degree 4, but it's much worse to write down; I won't
even try here.

There is no analogous formula for polynomials of degree
5. I don't just mean that no one has found the formula
yet; I mean that in 1826 Abel proved that there cannot
be such a formula. The problem is that the functions
don't do enough of what you need for
solving all 5th degree equations. (Imagine a calculator
that is missing a few buttons; there are some kinds of
calculations that you can't do on it.) You need at least
one more function. One such function, for instance, is
the inverse of the function f(x)=x5+x. (There are
other functions that would also work, and some of them
are more interesting to mathematicians for various
reasons, but I like this one because it can be described
in fairly elementary terms.) That function, together
with the functions
and addition, subtraction,
multiplication, and division is enough to give a formula
for the solution of the general 5th degree polynomial
equation in terms of the coefficients of the polynomial
- i.e., the degree 5 analogue of the quadratic formula.
But it's horribly complicated; I don't even want to think
about writing it down.